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    I'm stuck on this question and was wondering if anybody could provide some help.

    3) Log2x=8+9logx2

    I followed through my text book working and got an answer of 59049 instead of 0.5,512. Any help would be greatly appreciated ty
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    Could you make it clear what the bases are, is it to the base 10? x? 2?
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    (Original post by CapsLocke)
    Could you make it clear what the bases are, is it to the base 10? x? 2?
    My guess

    but of course there are other possibilities.
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    (Original post by steve2005)
    My guess

    but of course there are other possibilities.
    You're right bro,
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    You could start by changing the log-to-base-x into a log-to base-2
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    log_ab \equiv \frac{1}{log_ba}
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    This is my working:

    log2x=8+9logx2
    log2x=8+9/log2x

    -Let log2x = y

    y=8+9/y
    (x y)
    y^2=8y+9
    y^2-8y-9=0
    (y+1)(y-9)
    Y=-1 or 9

    Log2x=9
    x=9^2
    81

    or

    1

    whats wrong?
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    If log-to-base-2 (x) = 9
    x = 2^9
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    (Original post by ian.slater)
    If log-to-base-2 (x) = 9
    x = 2^9
    cheers bro
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    Not easy to put maths formulae. So I wrote the solution on a paper and uploaded the picture.
    Attached Images
     
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    You almost got the correct answer. log2 y=9 then y=2^9=512


    (Original post by joestevens2092)
    This is my working:

    log2x=8+9logx2
    log2x=8+9/log2x

    -Let log2x = y

    y=8+9/y
    (x y)
    y^2=8y+9
    y^2-8y-9=0
    (y+1)(y-9)
    Y=-1 or 9

    Log2x=9
    x=9^2
    81

    or

    1

    whats wrong?
 
 
 
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